by Stefano Maruelli –
Read & check, at your own risk !
Powers and Factorial: 
Updeated Probably due to the “Primes 1 How to discover if n is a prime 2 How to discover the position of any primes in the
primes list 3 How to find the next prime 1 How to discover if n is a prime: From primes definition I discover (but was probably
known from all mathematician from 1700d.c.) that using the “z” fraction: be “n” an
integer >= 5 if we calculate Rm: as
following, we found that (! =
Factorial):   an integer than n <> prime (1)
z = n!
/ n^2 =   
rational (from 24/5) than n = prime Unfortunately this method show in few numbers how .XLS
or similar non math programs are “limited” in the calculation of big numbers
like n!. Pari GP can help you
lot. What is interesting (that can be seen also in XLS
table till 17, only) is that: the non
integer part of z, in case of n=
prime start from 0.8 (or
24/5) and goes to 0.9999…. for z=
(infinite1) ! / infinite So prime has a very interesting
property that allow us to recognize them immediately, and count them
immediately (if we hold a supercomputer….) 1b Connection
to Wilson Theorem: The not so simple form of this function (called 2 How to
discover the position of any primes in the primes list With this simple trick you can understand the position
of the primes “n” in the primes table or how many primes there where before
the integer “n”: The official formula is: Where { X } is the non integer part of X, forced as 1 A more simple to understand method is:  Force to 0 the integer part of Rm  force at 1 the non integer part of the Rm value So in case “n” is a prime it count 1, or 0 in case
of non prime, so the sum from 1 to n will return exactly the number of the
primes. Since the method start from 5 we have to add 2 to
remember of: 2= prime and 3= prime, missed starting from 5: (0.3 pull decimal
to 1 in case n is a prime, 2 ) Here the LaTeX formula: {tex} Pi(x) = \sum_{n
= 5}^{\ P}{ [Int [[((n!/n^2)int(n!/n^2)]+0.3]]
+2 } {/tex} 3 How to find the next prime: With the similar method is possible to answer at the
question: If, for
example be Pi(a) = 31 is a known prime, witch is the
next prime ? The process is the same: 
calculate the position “i” of the known “Pi”
with the method (2) : so Pos(31) = i than Pos(Pi+1)=
(i+1) 
than knowing that the new position (i+1)
will be “easy” to 
calculate the relative prime One of the possible the tricks is:  Knowing that
P * 0 = 0 find a way to force
at zero any number that has a position different from (a+1) so first step is to calculate: This give as result:  0 if
n < P(i+1)  1 if n
= P(i+1)  K if n =
P(i+K) So we have to find a tricks that gives 0 or 1 still if n = P(i+K) and avoid
the indeterminate form 0/0. For example we know that b! = 1 still if b = 0 so:
int( b/b! ) avoid the form 0/0 And return 1 if
b=1 since if b =1
also b! = 1! = 1. So we use the: This give as result:  0 if
n <> P(i+1)  1 if n
= P(i+1) So to make it working itself we can put this trick
into a Sum that works from known limits where Pi(i+1)
will be for sure present. For example lower limit is: P(i)+1 and upper is:
2* Pi (as already proven see wikipedia) Of course the tricks works with the upper limit till
infinite, but has no sense. So the “final trick” to have the P(i+1) knowing Pi
is: All that works as a very slow computer program, so
has no sense for make a real calculation, but can give you an idea of what
make Primes soo hard to be discovered. So is necessary to “process” all the numbers from 5
to X each time, and for several times…) But we cannot say longer that “is impossible to find
a formula to calculate the next prime”. And finally we azzard to
say that seems now more probable that there will not be an absolutely easy
function that feet all primes. Of course there are other
more faster algorithm to find primes (for example Eartostene
method) but, in my opinion, they will not give a “sense” of how prime are
made as Wilkinson theorem (and what follow from it). There is non official formula discovered in 1964
that involves sin(x) and Integer operator too. 4 My final concerning on: I try to go over saying that is more clear now why
complex numbers can well fit the primes calculation: complex numbers, as primes, has 2 non connected “parts”: the real one and
the complex, as prime can be connected to a number z = n!/n^2
that has an integer part, that is common to a non primes
numbers, and a non integer part that is unique and non
present in non primes numbers. 
Solve the nroot by hand, and Fermat too ! A new story ? 

Power of integers properties to A possible way to Fermat proof using the Complicate Modulus Algebra THIS IS A WRONG METHOD ! 

How to easily understand integer solutions 

some concernings
on primes numbers 

Solution of ndegree
equations 

Fattorizzazione
“Repunit” (fattorizzazione
di numeri contenenti fattori del tipo “1….1”) 

Light Speed can be Doubled !!! Cern resuts confirm my
old “theory” ? Un nuovo modello “orbitale” tridimensionale 

